Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $p \neq 0$. $k = \dfrac{10p - 100}{p^2 - 5p - 50} \times \dfrac{3p^2 + 15p}{-2p + 18} $
First factor the quadratic. $k = \dfrac{10p - 100}{(p + 5)(p - 10)} \times \dfrac{3p^2 + 15p}{-2p + 18} $ Then factor out any other terms. $k = \dfrac{10(p - 10)}{(p + 5)(p - 10)} \times \dfrac{3p(p + 5)}{-2(p - 9)} $ Then multiply the two numerators and multiply the two denominators. $k = \dfrac{ 10(p - 10) \times 3p(p + 5) } { (p + 5)(p - 10) \times -2(p - 9) } $ $k = \dfrac{ 30p(p - 10)(p + 5)}{ -2(p + 5)(p - 10)(p - 9)} $ Notice that $(p - 10)$ and $(p + 5)$ appear in both the numerator and denominator so we can cancel them. $k = \dfrac{ 30p(p - 10)\cancel{(p + 5)}}{ -2\cancel{(p + 5)}(p - 10)(p - 9)} $ We are dividing by $p + 5$ , so $p + 5 \neq 0$ Therefore, $p \neq -5$ $k = \dfrac{ 30p\cancel{(p - 10)}\cancel{(p + 5)}}{ -2\cancel{(p + 5)}\cancel{(p - 10)}(p - 9)} $ We are dividing by $p - 10$ , so $p - 10 \neq 0$ Therefore, $p \neq 10$ $k = \dfrac{30p}{-2(p - 9)} $ $k = \dfrac{-15p}{p - 9} ; \space p \neq -5 ; \space p \neq 10 $